$A^*+A\geq0$ if and only if $A\geq0$

Question

Let H be a Hilbert space and $A:H\to H$ be a linear operator.

Definition (Positive) We say that A is positive (denoted by $A\geq0$) if $A=A^*$, i.e., $A$ is self-adjoint, and $$\langle Ax,x\rangle\geq0 \quad \forall x\in H.$$

Definition (Monotone) We say that A is monotone if $A^*+A\geq0$.

Lemma $A$ is monotone if and only if $\langle Ax,x\rangle\geq0$.

The above Lemma tells us that positive implies monotone, but not vice-versa because $A$ monotone does not guarantee $A$ self-adjoint.

Question 1: How to prove the above Lemma? (It should be trivial.)

1. $\langle Ax,x\rangle\geq0\Rightarrow A+A^*\geq0$.

From a routine computation $A+A^*\geq0$ if and only if $\mathrm{Re}\langle Ax,x\rangle\geq0$. Thus, it follows that if $\langle Ax,x\rangle\geq0$ then $\mathrm{Re}\langle Ax,x\rangle\geq0$.

2. $A+A^*\geq0\Rightarrow \langle Ax,x\rangle\geq0$. I am stuck here.

Answer 1

First, let's try to think first at the case of real Hilbert spaces. There, $A+A^*$ is positive iff $\langle(A+A^*)x,x\rangle\geq 0$ for every $x$, that is, $\langle Ax,x \rangle+\langle A^*x,x\rangle=\langle Ax,x \rangle+\langle x,Ax\rangle=2\langle Ax,x\rangle\geq 0 $, which is the definition of monotonicity (the last equality follows by the fact that the scalar product is symmetric, so $\langle Ax,x\rangle=\langle x,Ax\rangle$).

Now, in the complex case, the proof is just a modification of this argument. What changes is that now $\langle x,Ax\rangle=\overline{\langle Ax,x\rangle}$ since the scalar product is not symmetric anymore. This time $\langle Ax,x\rangle$ is not guaranteed to be a real number, but you make use of the fact that $$ \langle Ax,x\rangle+\overline{\langle Ax,x\rangle}=2\operatorname{Re}\{\langle Ax,x\rangle\}, $$ so in the end you obtain that $A+A^*$ is positive iff $\operatorname{Re}\{\langle Ax,x\rangle\}\geq 0$, which is as close as you can get to monotonicity if you have no additional hypotheses on the operator $A$ (again, $\langle Ax,x\rangle$ is not guaranteed to be a real number, so in general it doesn't make sense to check whether $\langle Ax,x\rangle\geq 0$).